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Thread: More questions on parametic curves in 2D

  1. #1
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    More questions on parametic curves in 2D

    Hi

    My question has two parts:
    Using the following two equation, $\displaystyle x=lnU$ & $\displaystyle y=ln(\frac{2}{U}$ find:
    1) calculate $\displaystyle \frac{d^2y}{dx^2}$

    This is what i have done:

    find $\displaystyle \frac{dy}{dx}$ first

    $\displaystyle \frac{dy}{dx} = \frac{dy}{dU} \frac{1}{\frac{dx}{dU}}$
    $\displaystyle \frac{dy}{dx} = \frac{-2U^{-2}}{2U^{-1}} \frac{1}{\frac{1}{U}}$
    $\displaystyle \frac{dy}{dx} = \frac{-u}{u}$
    $\displaystyle \frac{dy}{dx} = -1$

    $\displaystyle \frac{d^2y}{dx^2} = -U$

    where have a gone wrong?

    2) eliminate the parameter to give a direct relation between x and y:

    tried this way but was wrong.
    $\displaystyle e^x = U$

    $\displaystyle y=ln(\frac{2}{e^x}$

    answer in book is $\displaystyle y=ln2-x$
    P.S
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  2. #2
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    $\displaystyle \displaystyle \log\left( \frac{a}{b}\right)=\log a-\log b$,

    any base.
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  3. #3
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    Hello, Paymemoney!

    Given: .$\displaystyle \begin{Bmatrix} x &=& \ln u \\ y &=& \ln(\frac{2}{u}) \end{bmatrix}$

    1) Find: .$\displaystyle \dfrac{d^2y}{dx^2}$

    Note that: .$\displaystyle u > 0$

    We have: .$\displaystyle y \:=\:\ln2 - \ln u \quad\Rightarrow\quad \dfrac{dy}{du} \:=\:-\dfrac{1}{u}$

    And: .$\displaystyle x \:=\:\ln u \quad\Rightarrow\quad \dfrac{dx}{du} \:=\:\dfrac{1}{u}$


    Hence: .$\displaystyle \displaystyle \frac{dy}{dx} \;=\;\frac{\frac{dy}{du}}{\frac{dx}{du}} \;=\;\frac{\text{-}\frac{1}{u}}{\frac{1}{u}} \;=\;-1$

    Therefore: .$\displaystyle \dfrac{d^2y}{dx^2} \;=\;0$




    2) Eliminate the parameter to give a direct relation between $\displaystyle \,x$ and $\displaystyle \,y.$

    Answer in book is: $\displaystyle y\:=\:\ln2-x$

    We have:. . $\displaystyle \begin{array}{ccccccccc}
    x &=& \ln u & [1] \\
    y &=& \ln(\frac{2}{u}) & \Rightarrow & y &=& \ln 2 - \ln u & [2] \end{array}$


    Substitute [1] into [2]: .$\displaystyle y \;=\;\ln 2 - x$
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